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Power Series Operations · Tutorial 767 of 998

Integrating Products Through Convolution

Use convolution coefficients to build and check power-series antiderivatives of products within their common interval of convergence.

Intermediate 9 min read

What You'll Learn

  • State when the Cauchy product represents the product of two power series.
  • Compute product coefficients by summing over all index pairs with a fixed total degree.
  • Integrate the resulting power series term by term within its interval of convergence.
  • Include the integration constant and use the correct coefficient denominator.
  • Check a proposed antiderivative by differentiating its series.

From Product to Antiderivative

A product of two functions represented by power series can itself be represented by a power series. Its coefficients are not usually obtained by multiplying matching coefficients: each coefficient collects all the products of terms whose powers add to the same degree. This finite sum is called a convolution, or the Cauchy product coefficient.

Once the product coefficients are known, the termwise integration rule gives an antiderivative. Each coefficient is divided by the new power, and an arbitrary constant must be included. The order matters: first form the product coefficients, then integrate them.

Definition: Let \(f(x)=\sum_{n=0}^{\infty}a_n(x-a)^n\) and \(g(x)=\sum_{n=0}^{\infty}b_n(x-a)^n\) be power series with the same center \(a\). Their Cauchy product has coefficients $$ c_n=\sum_{k=0}^{n}a_k b_{n-k},\qquad n\ge 0. $$

For example, the coefficient \(c_2\) combines the three pairs of terms whose powers sum to \(2\): the constant term of one series with the quadratic term of the other, the two linear terms, and the quadratic term of the first with the constant term of the second.

$$ c_2=a_0b_2+a_1b_1+a_2b_0. $$

When the Product Series Is Valid

Suppose the two original series have radii of convergence \(R_f\) and \(R_g\), respectively. Write \(R=\min(R_f,R_g)\). For every \(x\) satisfying \(|x-a|<R\), both original series converge absolutely. On that interval their product is represented by the Cauchy product:

$$ f(x)g(x)=\sum_{n=0}^{\infty}c_n(x-a)^n, \qquad c_n=\sum_{k=0}^{n}a_kb_{n-k}. $$

The radius of convergence of this product series is at least \(R\). It may be larger: cancellation among the \(c_n\) can increase the radius. The guarantee that matters here is that the product expansion is valid throughout the common interior \(|x-a|<R\).

To see why the coefficient formula works, set \(t=x-a\). Multiplying the two series formally gives terms \(a_jb_k t^{j+k}\). Grouping together all terms with \(j+k=n\) produces exactly the finite sum defining \(c_n\). For \(|t|<R\), this grouping is legitimate because the double series of absolute values converges:

$$ \sum_{j=0}^{\infty}\sum_{k=0}^{\infty} |a_j|\,|b_k|\,|t|^{j+k} = \left(\sum_{j=0}^{\infty}|a_j||t|^j\right) \left(\sum_{k=0}^{\infty}|b_k||t|^k\right) <\infty. $$

Absolute convergence permits rearranging the terms by their total degree \(j+k\). This gives the Cauchy product expansion, not merely a formal multiplication rule.

Integrating the Product Coefficients

On an interval inside the common radius, the product is a continuous function represented by its power series. Integrating from the center \(a\) to \(x\) gives a convenient particular antiderivative. For any fixed \(x\) with \(|x-a|<R\), the segment between \(a\) and \(x\) lies inside the radius. On that segment the product series converges uniformly, so it may be integrated term by term.

Theorem: If \(f\) and \(g\) have power-series expansions centered at \(a\), with radii \(R_f\) and \(R_g\), then for \(|x-a|<\min(R_f,R_g)\), $$ \int_a^x f(t)g(t)\,dt = \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1}, \qquad c_n=\sum_{k=0}^{n}a_kb_{n-k}. $$ Every antiderivative of \(f(x)g(x)\) on this interval is the displayed series plus an arbitrary constant \(C\).

Here the integration variable is \(t\), while \(x\) is the endpoint. The coefficient \(c_n\) belongs to the power \((t-a)^n\). Its integral from \(a\) to \(x\) is \(c_n(x-a)^{n+1}/(n+1)\). The denominator is therefore the new exponent, \(n+1\), not the old exponent \(n\).

For justification, take partial sums of the product series. On the segment from \(a\) to \(x\), choose \(s\) with \(|x-a|\le s<R\). The series is uniformly convergent there, since it is absolutely bounded by the convergent series \(\sum_{n=0}^{\infty}|c_n|s^n\). Thus the integrals of the partial sums converge to the integral of the product. Integrating each finite partial sum yields the stated terms, and taking the limit gives the formula. Differentiating the resulting series inside the radius recovers the product series, so the expression is indeed an antiderivative. Any two antiderivatives on an interval differ by a constant.

Worked Example: Convolving Two Factorial-Coefficient Series

Worked Example: Convolving Two Factorial-Coefficient Series

Let \(f(x)=\sum_{n=0}^{\infty}x^n/n!\) and \(g(x)=\sum_{n=0}^{\infty}x^n/n!\). Both series converge for every real \(x\). Find a power series for an antiderivative of their product.

For a fixed \(n\), the product coefficient is

$$ c_n=\sum_{k=0}^{n}\frac{1}{k!}\frac{1}{(n-k)!} =\frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k} =\frac{2^n}{n!}. $$

The first coefficients are \(c_0=1\), \(c_1=2\), and \(c_2=2\). For instance, \(c_2=1/2!+1/(1!1!)+1/2!=1/2+1+1/2=2\), consistent with the formula.

Integrating from \(0\) to \(x\) gives the particular antiderivative

$$ F(x)=\sum_{n=0}^{\infty}\frac{2^n}{(n+1)n!}x^{n+1} =x+x^2+\frac{2}{3}x^3+\frac{1}{3}x^4+\cdots. $$

The displayed terms come from \(n=0,1,2,3\): their coefficients are \(1\), \(2/2=1\), \(4/(3\cdot2)=2/3\), and \(8/(4\cdot6)=1/3\). The full family of antiderivatives is \(F(x)+C\).

Check by differentiation. Termwise differentiation is valid at every real \(x\), since the series has infinite radius:

$$ F'(x)=\sum_{n=0}^{\infty} \frac{2^n}{(n+1)n!}(n+1)x^n =\sum_{n=0}^{\infty}\frac{2^n}{n!}x^n =\sum_{n=0}^{\infty}c_nx^n =f(x)g(x). $$

The derivative reproduces the product series coefficient by coefficient, as required.

Worked Example: A Product with Growing Coefficients

Worked Example: A Product with Growing Coefficients

For \(|x|<1\), let \(f(x)=\sum_{n=0}^{\infty}x^n\) and \(g(x)=\sum_{n=0}^{\infty}(n+1)x^n\). Compute the product coefficients and an antiderivative series.

For each \(n\), the first coefficient is \(a_k=1\), and the second is \(b_{n-k}=n-k+1\). Therefore

$$ c_n=\sum_{k=0}^{n}(n-k+1) =(n+1)+n+\cdots+1 =\frac{(n+1)(n+2)}{2}. $$

In particular, \(c_0=1\), \(c_1=3\), and \(c_2=6\). Directly, \(c_2=1\cdot3+1\cdot2+1\cdot1=6\).

Integrating each product term from \(0\) to \(x\) gives

$$ F(x)=\sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2(n+1)}x^{n+1} =\sum_{n=0}^{\infty}\frac{n+2}{2}x^{n+1} =x+\frac{3}{2}x^2+2x^3+\frac{5}{2}x^4+\cdots. $$

The first terms follow by substituting \(n=0,1,2,3\) into \((n+2)/2\). This series is valid for \(|x|<1\), the common interval of convergence. An arbitrary constant may be added.

Differentiate to check:

$$ F'(x)=\sum_{n=0}^{\infty}\frac{n+2}{2}(n+1)x^n =\sum_{n=0}^{\infty}\frac{(n+1)(n+2)}{2}x^n =\sum_{n=0}^{\infty}c_nx^n =f(x)g(x). $$

The coefficient of \(x^2\) in the derivative is \(6\), matching the convolution calculation.

Worked Example: Changing the Center

Worked Example: Changing the Center

Set \(t=x-2\), and consider \(f(x)=1+2t+3t^2\) and \(g(x)=\sum_{n=0}^{\infty}t^n/2^n\), which converges for \(|t|<2\). Find an antiderivative series centered at \(2\).

Write the polynomial coefficients as \(a_0=1\), \(a_1=2\), \(a_2=3\), and \(a_k=0\) for \(k>2\). The other coefficients are \(b_j=1/2^j\). The first product coefficients are \(c_0=1\) and \(c_1=b_1+2b_0=1/2+2=5/2\). For \(n\ge2\),

$$ c_n=b_n+2b_{n-1}+3b_{n-2} =\frac{1}{2^n}+\frac{2}{2^{n-1}}+\frac{3}{2^{n-2}} =\frac{17}{2^n}. $$

At \(n=2\), this gives \(c_2=17/4\); directly, \(c_2=b_2+2b_1+3b_0=1/4+1+3=17/4\).

Integrating with respect to \(t\), and using \(dt=dx\), gives

$$ F(x)=C+t+\frac{5}{4}t^2+ \sum_{n=2}^{\infty}\frac{17}{(n+1)2^n}t^{n+1}, \qquad t=x-2. $$

The first terms are \(C+t+\frac54t^2+\frac{17}{12}t^3+\frac{17}{32}t^4+\cdots\). The series is valid for \(|x-2|<2\). Differentiating gives \(1+\frac52t+\sum_{n=2}^{\infty}\frac{17}{2^n}t^n\), exactly the product series with coefficients \(c_0,c_1,c_n\) found above.

A Common Coefficient Error

A frequent mistake is to multiply coefficients with the same index and treat \(a_nb_n\) as the coefficient of degree \(n\). That ignores the other pairs of powers that contribute. For the series in the second example, the coefficient of \(x^2\) is not \(a_2b_2=1\cdot3=3\). The contributing pairs give

$$ c_2=a_0b_2+a_1b_1+a_2b_0 =1\cdot3+1\cdot2+1\cdot1 =6. $$

If the incorrect coefficient \(3\) were integrated, it would contribute \(x^3\), because \(3x^2\) integrates to \(x^3\). The correct product term \(6x^2\) contributes \(2x^3\). Differentiating the incorrect term gives \(3x^2\), not the product coefficient \(6x^2\), exposing the error.

A second useful check is the integration denominator. A term \(c_n(x-a)^n\) becomes \(c_n(x-a)^{n+1}/(n+1)\). Dividing by \(n\), or forgetting to add the integration constant when describing all antiderivatives, does not give the general antiderivative.

Why Convolution Is Useful

Convolution makes the effects of multiplication visible at the coefficient level. It is especially useful when a product function is easier to work with through known power series than through a direct integration technique. Rather than first simplifying the function, one can multiply the series, calculate the finite sum for each coefficient, and integrate the resulting series wherever both original expansions converge.

The procedure also provides a built-in check: after integrating, differentiation must return the product coefficients. If the derivative has a different coefficient at any degree, either the convolution or the integration step needs correction.

1
Check the common interval.
Work at points inside both original radii of convergence.
2
Form each product coefficient.
For degree \(n\), sum \(a_kb_{n-k}\) over \(k=0,\ldots,n\).
3
Integrate the product series.
Replace \(c_n(x-a)^n\) by \(c_n(x-a)^{n+1}/(n+1)\).
4
Check the result.
Differentiate the antiderivative series and compare its coefficients with the convolution.

Check Your Understanding

Use the coefficient and integration rules to answer the following questions.

  1. If \(c_3\) is a Cauchy product coefficient, which pairs of indices contribute to it?
  2. For \(a_n=1\) and \(b_n=n+1\), calculate \(c_2\) directly from the convolution sum.
  3. What coefficient and power result when \(c_n(x-a)^n\) is integrated?
  4. Why is the product-series formula guaranteed inside the smaller of the two radii?
  5. How can differentiating a proposed antiderivative detect an incorrect convolution coefficient?

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